## Sunday, January 3, 2010

### Problem 412: Triangle, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 412 about Triangle, Angles, Congruence.

Complete Problem 412
Level: High School, SAT Prep, College geometry

1. Another solution with full detail:
http://img9.imageshack.us/img9/8220/problem412.png

From C draw CE//AB and CE=AB ( see attached picture)
ABEC is a parallelogram and ∠ (ECy)= 2x
Triangle DEC is isosceles => ∠ (DEC)=1/2∠ (ECy)=x
Quadrilateral BDCE is cyclic with 2 opposite sides paralleled
So BDCE is an isosceles trapezoid
And ∠ (BDA)= ∠ (ECy)=2x => Triangles ABD is isosceles
In triangles ABD 4x=180-40 => x=35

2. My solution:
Draw circle 0 (circumcircle of triangle ABD)
Let E be the intersection of circle 0 with BC.
Construct AE and DE
Triangle BED is isosceles since ∠EBD =∠EAD = x
So ∠BAE = ∠BDE = x
Now ∠BDC =40º+2x so ∠EDC=40º+x since ∠BDE= x
Triangles ABE and EDC are similar by SAS
So ∠ DCE= x , finally x=35º since x+x+40º+2x=180º => 4x=140º => x=35º

1. http://i.imgur.com/K0x5Bvw.png

3. Let the angle bisector of A meet BC at E and let BE = p

Now < DAE = < DBE = x so ABED is cyclic

So (a-p)a = bc......(1)
Angie bisector theorem
p/(a-p) = c/b .......(2)

(1) X (2) gives us

ap = c^2 hence BC is tangential to Tr ACE at A and so < C = x

Now we know all the angles of Tr. ABC in terms of x and adding 4x+ 40 = 180 and x = 35

Sumith Peiris
Moratuwa
Sri Lanka

4. Or far more simply because ABED is cyclic BE = DE and Tr.s ABE and CDE are congruent SAS and so < C = x and the result follows

5. Problem 412
Brings BE=//DC then BECD is parallelogram so <BcE=x=< BAE=<BEA.Then BECA isisosceles trapezoid so <ACB=<AEB=x . Then 4x+40=180 so x=35.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

6. It is special case of x = 45-a/4 where a = 40.
The solution is same as Sumith Peiris.

7. See the drawing

Define point E in BC such as ∠DAE=x and ∠BAE=x
∠DAE=x and ∠DBE=x => ADEB are concyclic
∠BAE=x => ∠BDE=x
∠BDE=x and ∠BDE=x =>ΔBED is isosceles in E
∠BDC= ∠ABD+ ∠BAD= 40+2x and ∠BDE=x => ∠EDC=40+x
=> ∠EDC= ∠ABE= 40+x
=> ΔABE is congruent to ΔCDE (SAS)
=> EA=EC
=>ΔAEC is isosceles in E
=> ∠CAE=∠ACE=x
In ΔCDE : 2x+(40+x)+x= π
=>4x= π-40=140
Therefore x=35

8. In triangle ABD
sin(140-2x)/AB=sin2x/BD
AB/BD=sin(140-2x)/sin2x-------(1)

In triangle BCD
sinx/CD=sin(140-3x)/BD
CD/BD=sinx/sin(140-3x)-------(2)

As AB=CD, (1)=(2)
sin(140-2x)/sin2x=sinx/sin(140-3x)
sinxsin2x=sin(140-2x)sin(140-3x)
cosx-cos3x=cosx-cos(280-5x)
cos3x=cos(280-5x)
x=35